Le-Chaterlier principle and It’s application Questions in English

(D) According to Le Chatelier's principle,when an inert gas is added to a system at equilibrium at constant volume,the total pressure of the system increases,but the partial pressures of the reacting species remain unchanged.
Since the partial pressures (and thus concentrations) of the reactants and products do not change,the equilibrium position remains unaffected.
Therefore,the concentrations of $SO_2Cl_2$,$SO_2$,and $Cl_2$ will not change.

(D) When an inert gas like $Kr$ is added to a system at equilibrium at constant volume,the total pressure of the system increases,but the partial pressures of the reacting species remain unchanged.
Since the partial pressures of the reactants and products do not change,the reaction quotient $Q_c$ remains equal to the equilibrium constant $K_c$.
Therefore,the addition of an inert gas at constant volume has no effect on the equilibrium position,regardless of the value of $n$ (where $n = \Delta n_g$ is the change in the number of moles of gaseous species).

(C) According to Le Chatelier's principle:
$1$. For endothermic reactions $(\Delta H > 0)$,high temperature favors the forward reaction.
$2$. For exothermic reactions $(\Delta H < 0)$,low temperature favors the forward reaction.
$3$. For reactions where $\Delta n_g = 0$,the equilibrium is independent of pressure.
Analysis:
$P: A_{2(g)} B_{2(g)} \rightleftharpoons 2AB_{(g)}$ (Endothermic,$\Delta n_g = 0$). Favored by high temperature $(1)$ and is independent of pressure $(5)$.
$Q: 2AB_{2(g)} B_{2(g)} \rightleftharpoons 2AB_{3(g)}$ (Exothermic,$\Delta n_g = -1$). Favored by low temperature $(2)$.
$R: 2AB_{3(g)} \rightleftharpoons A_{2(g)} 3B_{2(g)}$ (Endothermic,$\Delta n_g = 2$). Favored by high temperature $(3)$.
Matching: $P-(1, 5), Q-(2), R-(3)$.
Correct option is $C$.

(C) The given reaction is $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$.
According to Le Chatelier's principle,the equilibrium shifts in the direction that counteracts the change.
Adding an inert gas at constant volume does not change the partial pressures of the reacting species,so the equilibrium remains unaffected.
Adding $Cl_2$ (a product) increases the concentration of the product,which shifts the equilibrium in the backward direction.
Adding an inert gas at constant pressure increases the total volume of the system.
Since the number of moles of gaseous products $(2)$ is greater than the number of moles of gaseous reactants $(1)$,increasing the volume shifts the equilibrium towards the side with more moles of gas,which is the forward direction.

(C) According to Le Chatelier's principle:
$1$. $NaNO_3$ and $NaNO_2$ are solids. Adding more solid does not change their active mass,so it does not shift the equilibrium.
$2$. The reaction involves the production of $O_{2(g)}$. Increasing the pressure will shift the equilibrium towards the side with fewer moles of gas,which is the backward direction (towards $NaNO_3$).
$3$. The decomposition of $NaNO_3$ is an endothermic process $(\Delta H > 0)$. According to Le Chatelier's principle,decreasing the temperature will shift the equilibrium in the backward direction (exothermic direction).
$4$. Therefore,option $C$ is correct because increasing pressure favors the side with fewer gaseous moles.

(A) According to Le Chatelier's principle:
$1$. For an exothermic reaction $(\Delta H < 0)$,a decrease in temperature favors the forward reaction to produce more product.
$2$. The reaction involves a change in the number of moles of gas: $\Delta n_g = 2 - (1 + 4) = 2 - 5 = -3$.
$3$. Since $\Delta n_g < 0$,an increase in pressure shifts the equilibrium towards the side with fewer moles of gas,which is the product side.
$4$. Therefore,low temperature and high pressure favor the formation of the product $XY_4$.

(C) $1$. Consider the two equilibrium reactions occurring in the same container:
$(i)$ $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$
(ii) $COCl_{2(g)} \rightleftharpoons CO_{(g)} + Cl_{2(g)}$
$2$. When $CO_{(g)}$ is added to the container,according to Le Chatelier's principle,the equilibrium of reaction (ii) will shift in the backward direction to consume the added $CO_{(g)}$.
$3$. As reaction (ii) shifts backward,the concentration of $Cl_{2(g)}$ decreases.
$4$. Now,consider reaction $(i)$. Since the concentration of $Cl_{2(g)}$ has decreased,the equilibrium of reaction $(i)$ will shift in the forward direction to compensate for the loss of $Cl_{2(g)}$.
$5$. As reaction $(i)$ shifts forward,$PCl_{5(g)}$ dissociates to form more $PCl_{3(g)}$ and $Cl_{2(g)}$.
$6$. Therefore,the amount of $PCl_{5(g)}$ decreases.

(B) $1$. The given reaction is $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} + 21.9 \, kcal$.
$2$. The reaction is exothermic $(\Delta H < 0)$,so according to Le Chatelier's principle,a low temperature favors the forward reaction (production of $NH_3$).
$3$. The number of moles of gaseous reactants is $1 + 3 = 4$,and the number of moles of gaseous products is $2$. Since the forward reaction involves a decrease in the number of moles,high pressure favors the forward reaction.
$4$. Therefore,low temperature and high pressure are the ideal conditions for the production of ammonia.

(C) According to Le Chatelier's principle:
$1$. For high temperature: The reaction must be endothermic $(\Delta H = +ve)$ so that heat is absorbed to shift the equilibrium forward.
$2$. For high pressure: The reaction must result in a decrease in the number of moles of gaseous products compared to reactants $(\Delta n_g < 0)$.
$3$. Analyzing option $C$: $Cl_{2(g)} + 2O_{2(g)} \rightleftharpoons 2ClO_{2(g)}$. Here,$\Delta H = +ve$ (endothermic) and $\Delta n_g = 2 - (1+2) = -1$. Since $\Delta n_g < 0$,high pressure favors the forward reaction. Thus,both high temperature and high pressure favor the product yield in reaction $C$.

(C) The given reaction is $Cl_{2(g)} + 3F_{2(g)} \rightleftharpoons 2ClF_{3(g)}$ with $\Delta H = -329 \, kJ$.
Since $\Delta H < 0$,the reaction is exothermic.
According to Le Chatelier's principle,for an exothermic reaction,increasing the temperature shifts the equilibrium to the left (reactant side),decreasing the yield of $ClF_3$.
Adding more reactant $(F_2)$ shifts the equilibrium to the right (product side),increasing the amount of $ClF_3$.
Removing $Cl_2$ shifts the equilibrium to the left,decreasing the yield.
Increasing the volume of the container favors the side with more moles of gas. Here,reactants have $1 + 3 = 4$ moles and products have $2$ moles. Increasing volume shifts the equilibrium to the left,decreasing the yield of $ClF_3$.

(A) According to Le Chatelier's principle,an increase in pressure shifts the equilibrium towards the side with a smaller number of moles of gaseous species.
For reaction $A$: $\Delta n_g = (1 + 1) - 1 = +1$. Since the number of moles of products is greater than reactants,increasing pressure shifts the equilibrium in the backward direction.
For reactions $B$ and $C$,$\Delta n_g$ is negative,so increasing pressure shifts them in the forward direction.
For reaction $D$,$\Delta n_g = 2 - (1 + 1) = 0$,so pressure has no effect.

(A) For the reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,the number of moles of gaseous products $(2 \ mol)$ is greater than the number of moles of gaseous reactants $(1 \ mol)$.
According to Le Chatelier's principle,increasing the pressure (or compressing the mixture) shifts the equilibrium toward the side with fewer moles of gas to reduce the pressure.
Therefore,the equilibrium shifts in the backward direction (toward the reactant side).

(C) The given reaction is $CO_{2(s)} \rightleftharpoons CO_{2(g)}$,$\Delta H = +ve$.
Since $\Delta H$ is positive,the reaction is endothermic.
According to Le Chatelier's principle,for an endothermic reaction,a decrease in temperature shifts the equilibrium in the backward direction.
Therefore,the concentration of the reactant,$CO_{2(s)}$,increases.

(C) According to Le Chatelier's principle,an increase in temperature favors the endothermic reaction (where $\Delta H > 0$).
$1$. The reaction $CO_{(g)} + H_2O_{(g)} \rightleftharpoons CO_{2(g)} + H_{2(g)}$ is exothermic.
$2$. The reaction $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$ is exothermic.
$3$. The reaction $H_2O_{(g)} \rightleftharpoons H_{2(g)} + \frac{1}{2}O_{2(g)}$ is endothermic $(\Delta H > 0)$.
$4$. The reaction $4HCl_{(g)} + O_{2(g)} \rightleftharpoons 2H_2O_{(g)} + 2Cl_{2(g)}$ is exothermic.
Therefore,increasing the temperature will shift the equilibrium in the forward direction for the endothermic reaction,which is option $C$.

(C) The given reaction is $2A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)} + 362 \, kcal$.
Since the reaction is exothermic $(\Delta H < 0)$,according to Le Chatelier's principle,a lower temperature will shift the equilibrium to the right,favoring the formation of products.
For the pressure effect,we look at the change in the number of moles of gas: $\Delta n_g = n_{\text{products}} - n_{\text{reactants}} = 1 - (2 + 1) = -2$.
Since $\Delta n_g < 0$,an increase in pressure will shift the equilibrium towards the side with fewer moles of gas,which is the product side.
Therefore,high pressure and low temperature favor the formation of $C$.

(A) Le Chatelier's principle states that if a system at equilibrium is subjected to a change in concentration,temperature,or pressure,the system will adjust itself to counteract the effect of the change and establish a new equilibrium. This principle is applicable only to $A$ (Reversible reactions) that are in a state of dynamic equilibrium.

(B) According to Le Chatelier's principle,an increase in pressure shifts the equilibrium towards the side with fewer moles of gaseous species.
In the given reaction,the number of moles of gaseous reactants is $1$ $(H_2O)$ and the number of moles of gaseous products is $2$ $(H_2 + CO)$.
Since the number of moles of gaseous products is greater than the gaseous reactants,an increase in pressure will shift the equilibrium towards the side with fewer moles,which is the backward direction.

(A) According to Le Chatelier's principle,adding more $SCN^-$ ions to the equilibrium mixture will shift the reaction in the forward direction to consume the added reactant.
This results in an increased concentration of the product $[Fe(SCN)]^{2+}$.
Therefore,the intensity of the deep red color will increase.

(D) According to Le Chatelier's principle,if the volume of the reaction vessel is increased,the pressure decreases.
To counteract this,the equilibrium shifts towards the side with a greater number of moles of gas.
Given that the degree of dissociation decreases when the volume increases,it implies that the equilibrium shifts in the backward direction (towards the reactant $A$).
This happens when the number of moles of gaseous products is less than the number of moles of gaseous reactants.
Therefore,$(l + m) < a$ or $a > (l + m)$.

(C) The given reaction is $H_2 \rightleftharpoons H + H$.
Since $\Delta H = +ve$,the reaction is endothermic.
According to Le Chatelier's principle,an increase in temperature favors the endothermic direction.
Also,the number of moles of gas increases from $1$ to $2$ in the forward reaction.
Therefore,a decrease in pressure favors the forward reaction.
Thus,high temperature and low pressure are favorable conditions.

(B) The reaction is $2NO \rightleftharpoons N_2 + O_2$ (exothermic).
According to Le Chatelier's principle,for an exothermic reaction,a decrease in temperature shifts the equilibrium in the forward direction.
Therefore,the dissociation of $NO$ increases with a decrease in temperature.

(D) The given reaction is an exothermic reaction: $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) + \text{Heat}$.
According to Le Chatelier's principle,for an exothermic reaction,decreasing the temperature favors the forward reaction.
Additionally,increasing the concentration of reactants (like $O_2$ or $SO_2$) or decreasing the concentration of products (like $SO_3$) shifts the equilibrium in the forward direction.
Among the given options,adding oxygen $(O_2)$ increases the concentration of reactants,which shifts the equilibrium to the right (forward direction).

(A) For a reversible reaction,the relationship between $K_p$ and $K_c$ is $K_p = K_c(RT)^{\Delta n}$.
Given $K_c > K_p$,it implies that $(RT)^{\Delta n} < 1$,which means $\Delta n$ must be negative.
$A$ negative $\Delta n$ indicates that the number of moles of gaseous products is less than the number of moles of gaseous reactants.
According to Le Chatelier's principle,an increase in pressure shifts the equilibrium towards the side with fewer moles,favoring the product.
Conversely,a decrease in pressure will shift the equilibrium towards the reactant side,resulting in less product formation.
Given $\Delta H = +40 \, kcal$,the reaction is endothermic.
According to Le Chatelier's principle,an increase in temperature favors the endothermic direction (product side).
Conversely,a decrease in temperature will shift the equilibrium towards the reactant side,resulting in less product formation.
Therefore,a decrease in both pressure and temperature will result in less product formation.

(A) The given reaction is $3X_{(g)} + Y_{(g)} \rightleftharpoons X_3Y_{(g)}$.
According to Le Chatelier's principle,the equilibrium position is affected by changes in concentration,pressure,and temperature.
$1$. Pressure: Since the number of moles of gaseous reactants $(3+1=4)$ is different from the number of moles of gaseous products $(1)$,a change in pressure will shift the equilibrium.
$2$. Temperature: The formation of $X_3Y$ from $X$ and $Y$ involves a change in enthalpy,so temperature will affect the equilibrium constant and the position of equilibrium.
$3$. Catalyst: $A$ catalyst only increases the rate of both forward and backward reactions equally and does not change the equilibrium position or the amount of products at equilibrium.
Therefore,the amount of $X_3Y_{(g)}$ is affected by both temperature and pressure.

(C) According to Le Chatelier's principle,when an inert gas is added to a reaction at constant pressure,the volume of the system increases to maintain the pressure.
This leads to a decrease in the partial pressure of the reacting species.
The equilibrium shifts in the direction where the total number of moles of gaseous products is greater.
For the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the number of moles of gaseous reactants is $1 + 3 = 4$,and the number of moles of gaseous products is $2$.
Since the number of moles of gaseous reactants $(4)$ is greater than the number of moles of gaseous products $(2)$,the equilibrium will shift towards the side with more moles to compensate for the volume increase.
Therefore,the equilibrium shifts towards the left (towards the reactants).

(B) The given reaction is $2NO \rightleftharpoons N_2 + O_2 + x \, cal$.
Since the reaction releases heat $(x \, cal)$,it is an exothermic reaction.
According to Le Chatelier's principle,for an exothermic reaction,a decrease in temperature favors the forward reaction.
However,the question asks for the dissociation of $NO$. Since the number of moles of gaseous reactants ($2$ moles) is equal to the number of moles of gaseous products ($1+1=2$ moles),pressure has no effect on the equilibrium.
Therefore,to favor the forward reaction (dissociation of $NO$),a low temperature is required.

(A) According to Le Chatelier's principle,if a product $(Cl_2)$ is added to the system at equilibrium,the equilibrium will shift in the backward direction to counteract the change.
$(1)$ The addition of $Cl_2$ leads to a change in the concentrations of all species ($SO_2$,$Cl_2$,and $SO_2Cl_2$) as the system re-establishes equilibrium.
$(2)$ Since the reaction shifts backward,$Cl_2$ is consumed,not produced. Thus,statement $(2)$ is incorrect.
$(3)$ As the reaction shifts backward,$SO_2$ reacts with $Cl_2$ to form more $SO_2Cl_2$. Therefore,the concentration of $SO_2$ decreases and the concentration of $SO_2Cl_2$ increases. Thus,statement $(3)$ is correct.
Therefore,statements $(1)$ and $(3)$ are correct.

(B) According to Le Chatelier's principle,if the pressure on a gaseous system at equilibrium is decreased,the equilibrium shifts in the direction where the number of moles of gaseous species increases.
In the given reaction: $2A_{(s)} + 3B_{(g)} \rightleftharpoons 3C_{(g)} + D_{(g)} + Q$.
The number of moles of gaseous reactants is $3$ (from $3B_{(g)}$).
The number of moles of gaseous products is $3 + 1 = 4$ (from $3C_{(g)} + D_{(g)}$).
Since the number of moles of gaseous products $(4)$ is greater than the number of moles of gaseous reactants $(3)$,a decrease in pressure will shift the equilibrium towards the product side (right side).
Therefore,the concentrations of products $C$ and $D$ will increase.

(C) The dissociation of molecular hydrogen into atomic hydrogen is represented by the equation: $H_{2(g)} \rightleftharpoons 2H_{(g)}$.
This reaction involves the breaking of a chemical bond,which is an endothermic process $(\Delta H > 0)$.
According to Le Chatelier's principle,for an endothermic reaction,an increase in temperature favors the forward reaction.
Furthermore,the number of moles of gas increases from $1 \text{ mole}$ on the reactant side to $2 \text{ moles}$ on the product side.
According to Le Chatelier's principle,a decrease in pressure favors the side with a greater number of moles of gas.
Therefore,high temperature and low pressure are the ideal conditions for the dissociation of $H_2$ into $H$.

(A) For the reaction $Br_2 \rightleftharpoons 2Br$,the equilibrium constants are given as:
At $T_1 = 500 \, K$,$K_1 = 1 \times 10^{-10}$
At $T_2 = 700 \, K$,$K_2 = 1 \times 10^{-5}$
Since the temperature increases $(T_2 > T_1)$ and the equilibrium constant also increases $(K_2 > K_1)$,the reaction is endothermic.
According to Le Chatelier's principle,for an endothermic reaction,an increase in temperature shifts the equilibrium in the forward direction,leading to a higher value of the equilibrium constant $K$.

(D) According to Le Chatelier's principle:
$1$. For the equilibrium to shift in the backward direction at low pressure,the number of moles of gaseous products must be less than the number of moles of gaseous reactants,i.e.,$(a + b) > (c + d)$.
$2$. For the equilibrium to shift in the backward direction at high temperature,the forward reaction must be exothermic,i.e.,$\Delta H < 0$.

(D) According to $Le \text{ } Chatelier's$ principle,for a reaction at equilibrium,an increase in pressure favors the direction in which the total number of moles of gas decreases.
In the reaction $CO_{(g)} + 3H_{2(g)} \rightleftharpoons CH_{4(g)} + H_2O_{(g)}$,the number of moles of gaseous reactants is $1 + 3 = 4$,and the number of moles of gaseous products is $1 + 1 = 2$.
Since the number of moles decreases in the forward direction $(4 \rightarrow 2)$,an increase in pressure will shift the equilibrium to the right,thereby increasing the yield of products.

(B) According to $Le \ Chatelier's \ principle$,if the pressure is increased on a system at equilibrium,the system will shift in the direction that reduces the number of moles of gas.
In the reaction $H_2O_{(l)} \rightleftharpoons H_2O_{(g)}$,the product side has $1 \ mole$ of gas,while the reactant side has $0 \ moles$ of gas.
Increasing the pressure favors the formation of the liquid phase (the side with fewer moles of gas).
Therefore,the equilibrium shifts in the backward direction.

(A) According to Le Chatelier's principle,a change in volume affects the equilibrium position only if there is a change in the total number of moles of gaseous species,i.e.,$\Delta n_g \neq 0$.
For option $A$: $\Delta n_g = (2) - (1 + 1) = 0$.
Since $\Delta n_g = 0$,a change in volume will not shift the equilibrium,and the number of moles of components will remain unchanged.
For options $B$,$C$,and $D$,$\Delta n_g \neq 0$,so a change in volume will affect the equilibrium.

(A) In the equilibrium $Ice \rightleftharpoons Water$,the density of water is higher than that of ice,meaning the volume decreases during the melting process. According to Le Chatelier's principle,an increase in pressure favors the direction that results in a decrease in volume. Therefore,increasing the pressure shifts the equilibrium towards the formation of water,which effectively lowers the melting point of ice.

(C) According to Le Chatelier's principle,when the total pressure of a system at equilibrium is decreased,the equilibrium shifts in the direction where the number of moles of gaseous species increases.
For option $A$: $\Delta n_g = 2 - (1 + 3) = -2$ (Decreasing pressure shifts equilibrium to the left).
For option $B$: $\Delta n_g = 2 - (1 + 1) = 0$ (Pressure change has no effect).
For option $C$: $\Delta n_g = 2 - 1 = +1$ (Decreasing pressure shifts equilibrium to the right).
For option $D$: $\Delta n_g = 2 - (1 + 1) = 0$ (Pressure change has no effect).
Therefore,the correct option is $C$.

(C) According to Le Chatelier's principle:
$(1)$ Adding an inert gas at constant volume does not change the partial pressures of the reactants or products,so the equilibrium remains unaffected.
$(2)$ Adding $Cl_{2(g)}$ at constant volume increases the concentration of the product,which shifts the equilibrium in the backward direction.
$(3)$ Adding an inert gas at constant pressure increases the total volume of the system,which decreases the partial pressure of each gas. Since the number of moles of products $(2)$ is greater than the number of moles of reactants $(1)$,the equilibrium shifts in the forward direction.
$(4)$ Increasing the volume of the container decreases the total pressure. According to Le Chatelier's principle,the equilibrium shifts towards the side with more moles of gas,which is the product side (forward direction).
$(5)$ Adding $PCl_{5(g)}$ at constant volume increases the concentration of the reactant,which shifts the equilibrium in the forward direction.
Therefore,the forward reaction is favored by $(3), (4),$ and $(5)$.

(B) The chemical equilibrium reaction is: $2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g)$.
According to Le Chatelier's principle,if the total pressure of a system at equilibrium is increased at constant temperature and volume,the equilibrium will shift in the direction that results in a decrease in the number of moles of gas.
In this reaction,the reactant side has $2$ moles of gas,and the product side has $2 + 1 = 3$ moles of gas.
Increasing the total pressure will shift the equilibrium towards the side with fewer moles of gas,which is the reactant side $(SO_3)$.
Therefore,the dissociation of $SO_3$ will decrease.

(B) According to Le Chatelier's principle,for a reaction at equilibrium,an increase in pressure shifts the equilibrium towards the side with a smaller number of moles of gaseous species.
For reaction $A$: $\Delta n_g = (1+1) - 1 = 1$. Increasing pressure shifts equilibrium to the left.
For reaction $B$: $\Delta n_g = 2 - (1+1) = 0$. The number of moles of gaseous reactants and products is equal,so pressure change has no effect on the yield.
For reaction $C$: $\Delta n_g = 2 - (1+3) = -2$. Increasing pressure shifts equilibrium to the right (yield increases).
For reaction $D$: $\Delta n_g = 2 - (2+1) = -1$. Increasing pressure shifts equilibrium to the right (yield increases).
Therefore,in reaction $B$,the yield does not increase with high pressure.

(A) According to Le Chatelier's principle,increasing the pressure favors the direction in which the volume decreases (or density increases).
Density is defined as $\text{Density} = \frac{\text{Mass}}{\text{Volume}}$,which implies $\text{Volume} = \frac{\text{Mass}}{\text{Density}}$.
Since the density of diamond $(3.5 \ g/mL)$ is greater than that of graphite $(2.3 \ g/mL)$,the volume of $1 \ g$ of diamond is less than the volume of $1 \ g$ of graphite.
Therefore,increasing the pressure shifts the equilibrium towards the side with the smaller volume,which is the diamond side.
Thus,the backward reaction is favored.

(B) The given reaction is $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) + \text{Heat}$.
According to Le Chatelier's principle,for an exothermic reaction,a decrease in temperature shifts the equilibrium in the forward direction to release more heat,thus increasing the yield of $SO_3$.
Since the number of moles of gaseous products $(2 \ mol)$ is less than the number of moles of gaseous reactants $(2+1 = 3 \ mol)$,an increase in pressure shifts the equilibrium towards the side with fewer moles (forward direction) to reduce the pressure.
Therefore,decreasing the temperature and increasing the pressure will increase the yield of $SO_3$.

(A) The given reaction is $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$ with $\Delta H^o = -198 \, kJ$.
Since $\Delta H^o < 0$,the reaction is exothermic.
According to Le Chatelier's principle,for an exothermic reaction,a decrease in temperature favors the forward reaction.
For the pressure effect,we look at the change in the number of moles of gas: $\Delta n_g = 2 - (2 + 1) = -1$.
Since the number of moles decreases in the forward direction,an increase in pressure favors the forward reaction.
Therefore,the favorable conditions are a decrease in temperature and an increase in pressure.

(B) The equilibrium between dichromate and chromate ions is given by: $Cr_2O_7^{2-} + 2OH^- \rightleftharpoons 2CrO_4^{2-} + H_2O$.
In a basic medium,the concentration of $OH^-$ ions increases.
According to Le Chatelier's principle,adding a reactant $(OH^-)$ shifts the equilibrium in the forward direction (to the right) to produce more $CrO_4^{2-}$ ions.
Therefore,the equilibrium shifts to the right in a basic medium.

(D) catalyst provides an alternative reaction pathway with a lower activation energy $(E_a)$.
In a reversible reaction,the catalyst lowers the activation energy for both the forward and the backward reactions by the same amount.
Therefore,it increases the rate of both the forward and backward reactions to the same extent,allowing the system to reach equilibrium faster without changing the equilibrium constant $(K_{eq})$.

(C) catalyst in a reversible reaction provides an alternative pathway with lower activation energy for both the forward and backward reactions. As a result,it increases the rate of both reactions equally,thereby decreasing the time required to reach the state of equilibrium.

(B) The given reaction is $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$ with $\Delta H = -45.0 \ kcal$.
Since the reaction is exothermic $(\Delta H < 0)$,according to Le Chatelier's principle,an increase in temperature will shift the equilibrium in the backward direction to absorb the excess heat.
Therefore,high temperature is not favourable for the formation of $SO_3$.